Question

Two electric charge +q and+4q are placed at a distance 6a apart on a horizontal plane. Find the position of the point on the line joining the two charges where the electric field is zero

Answer 1

Answer:

Given:

+4q and +q charges are placed at distance of 6a from each other.

To find:

Position on the line where electric field is zero.

Calculation:

Let that position be at distance x from charge q . So distance from 4q will be(6a - x).

$\dfrac{1}{4\pi \epsilon} \dfrac{q}{ {x}^{2} } = \dfrac{1}{4\pi \epsilon} \dfrac{4q}{ {(6a - x)}^{2} }$

$= > \dfrac{1}{ {x}^{2} } = \dfrac{4}{ {(6a - x)}^{2} }$

Square root on both sides :

$= > \dfrac{1}{ x } = \dfrac{2}{ (6a - x)}$

$= > 6a - x = 2x$

$= > 3x = 6a$

$= > x = 2a$

So the point is 2a unit from q charge and 4a from 4q charge.

Answer 2

$\underline{\boxed{\mathfrak{ \huge{ \purple{Answer}}}}}$

Given :

Two electrical point charge of +q and +4q are placed at distance 6a apart on the horizontal plane.

To Find :

Point where net electric field due to both charges become zero.

Assumption :

Let x is the distance from charge +q and (6a-x) is the distance from charge +4q where net electric field become zero.

Electric field at that point due to +q charge is E1 and due to +4q charge is E2.

E1 = E2 is the condition for zero electric field.

Formula :

Electric field due to a point charge at r distance is given by...

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \dag \: \: \underline{ \boxed{ \bold{ \pink{ \rm{E = \frac{kq}{ {r}^{2} } }}}}} \: \: \dag$

Calculation :

$\implies \rm \: E_{1} = E_{2} \\ \\ \therefore \rm \: \frac{kq}{ {x}^{2} } = \frac{4kq}{ {(6a - x)}^{2} } \\ \\ \therefore \rm \: \frac{6a - x}{x} = \sqrt{4} = 2 \\ \\ \therefore \rm \: 6a - x = 2x \\ \\ \therefore \rm \: 6a = 3x \\ \\ \therefore \rm \: \boxed{ \rm{\red{x = 2a}}}$

A point which is situated at 2a distance from +q charge and 4a distance from +4q charge net electric field is zero.